Question

How do you solve using the quadratic formula for `y = 12x^2 - 35x + 8`?

Answer 1

`y = 12x^2 - 35x + 8 = 0`

`D = d^2 = b^2 - 4ac = 1225 - 384 = 841 -> d = +- 29`

`x = 35/24 + 29/24 = 64/24 = 8/3`

`x = 35/24 - 29/24 = 6/24 = 1/4`

There is another way: solving by the new Transforming Method (Google , Yahoo Search) that may be faster and not boring, especially when you can't use calculator.
`y = 12x^2 - 35x + 8`
Transformed equation : `y' = x^2 - 35x + 96 (a.c = 12(8) = 96)`
Solve y' by composing factor pairs of (96). Proceed: (2, 48)(3, 32).
This last sum is 35 = -b. Then its 2 real roots are: p' = 3 and q' = 32. Back to original equation, the 2 real roots are: `p = (p')/a = 3/12 = 1/4, and q = (q')/a = 32/12 = 8/3.`